What is the integral of (x+1)/(4x^2-4x+1) ?

The integral of (x+1)/(4x^2-4x+1) is 1/4 ln|2x-1|-3/(4(2x-1))+C

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integral of (x+1)/(4x^2-4x+1)

\int \frac{x + 1}{4 x ^{2} - 4 x + 1} d x

\frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )} + C

Solution steps

\int \frac{x + 1}{4 x ^{2} - 4 x + 1} d x

Take the partial fraction of

\frac{x + 1}{4 x ^{2} - 4 x + 1} : \frac{1}{2 ( 2 x - 1 )} + \frac{3}{2 ( 2 x - 1 ) ^{2}}

= \int \frac{1}{2 ( 2 x - 1 )} + \frac{3}{2 ( 2 x - 1 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{2 ( 2 x - 1 )} d x + \int \frac{3}{2 ( 2 x - 1 ) ^{2}} d x

\int \frac{1}{2 ( 2 x - 1 )} d x = \frac{1}{4} ln ∣ 2 x - 1 ∣

\int \frac{3}{2 ( 2 x - 1 ) ^{2}} d x = - \frac{3}{4 ( 2 x - 1 )}

= \frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )}

Add a constant to the solution

= \frac{1}{4} ln ∣ 2 x - 1 ∣ - \frac{3}{4 ( 2 x - 1 )} + C

What is the integral of (x+1)/(4x^2-4x+1) ?
The integral of (x+1)/(4x^2-4x+1) is 1/4 ln|2x-1|-3/(4(2x-1))+C